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प्रश्न
f(x) = \[x\sqrt{2 - x^2} - \sqrt{2} \leq x \leq \sqrt{2}\] .
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उत्तर
\[\text { Given }: f\left( x \right) = x\sqrt{2 - x^2}\]
\[ \Rightarrow f'\left( x \right) = \sqrt{2 - x^2} - \frac{x^2}{\sqrt{2 - x^2}}\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \sqrt{2 - x^2} - \frac{x^2}{\sqrt{2 - x^2}} = 0\]
\[ \Rightarrow \sqrt{2 - x^2} = \frac{x}{\sqrt{2 - x^2}}\]
\[ \Rightarrow 2 - x^2 = x^2 \]
\[ \Rightarrow x^2 = 1\]
\[ \Rightarrow x = \pm 1 \]
\[\text { Thus, x = 1 and x = - 1 are the possible points of local maxima or local minima }. \]
\[\text { Now }, \]
\[f''\left( x \right) = \frac{- x}{\sqrt{2 - x^2}} - \left( \frac{2x\sqrt{2 - x^2} + \frac{x^3}{\sqrt{2 - x^2}}}{2 - x^2} \right) = \frac{- x}{\sqrt{2 - x^2}} - \left( \frac{2x\left( 2 - x^2 \right) + x^3}{\left( 2 - x^2 \right)\sqrt{2 - x^2}} \right)\]
\[\text { At }x = 1: \]
\[ f''\left( 1 \right) = \frac{- 1}{\sqrt{2 - 1^2}} - \left[ \frac{2\left( 2 - 1^2 \right) + 1^3}{\left( 2 - 1^2 \right)\sqrt{2 - 1^2}} \right] = - \frac{1}{2} - \frac{3}{2} = - 2 < 0\]
\[\text { So, x = 1 is the point of local maximum }. \]
\[\text { The local maximum value is given by }\]
\[f\left( 4 \right) = 1\sqrt{2 - 1^2} = 1\]
\[\text { At }x = - 1: \]
\[ f''\left( - 1 \right) = \frac{1}{\sqrt{2 - 1^2}} + \left[ \frac{2\left( 2 - 1^2 \right) - 1^3}{\left( 2 - 1^2 \right)\sqrt{2 - 1^2}} \right] = 1 + 1 = 2 > 0\]
\[\text { So, x = - 1 is the point of local minimum } . \]
\[\text { The local minimum value is given by }\]
\[f\left( - 1 \right) = - 1\sqrt{2 - 1^2} = - 1\]
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